SOLUTION OF HW8 November 06, 2011 1. Show that lim

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SOLUTION OF HW8
MINGFENG ZHAO
November 06, 2011
1. Show that
lim f (x) = A if and only if
x→p
lim [f (x) − A] = 0.
x→p
Proof. (=⇒) If lim f (x) = A, then for any > 0, there exists some δ > 0 such that
x→p
|f (x) − A| < ,
for all |x − p| < δ.
That is, lim [f (x) − A] = 0.
x→p
(⇐=) If lim [f (x) − A] = 0, then for any > 0, there exists some δ > 0 such that
x→p
|f (x) − A| < ,
for all |x − p| < δ.
That is, lim f (x) = A.
x→p
2. If lim f (x) = A, and lim g(x) = B, show that
x→p
x→p
lim [f (x) − g(x)] = A − B.
x→p
Proof. Since lim f (x) = A, then for any > 0, there exists some δ1 > 0 such that
x→p
|f (x) − A| <
,
2
for all |x − p| < δ1 .
For the above > 0, since lim g(x) = B, then there exists some δ2 > 0 such that
x→p
|g(x) − B| <
,
2
for all |x − p| < δ2 .
1
2
MINGFENG ZHAO
Now taking δ = min {δ1 , δ2 } > 0, for all |x − p| < δ, we have
|f (x) − A| <
,
2
and |g(x) − B| <
.
2
Then we have
|[f (x) − g(x)] − [A − B]| = |f (x) − A + g(x) − B|
≤
|f (x) − A| + |g(x) − B|
<
+
2 2
= ,
for all |x − p| < δ.
That is, lim [f (x) − g(x)] = A − B.
x→p
3. Show that lim x3 = p3 for all p ∈ R
x→p
Proof. For any p ∈ R and fix, for any |x−p| < 1, we know that |x| = |x−p+p| ≤ |x−p|+|p| < 1+|p|.
For any > 0, we know that for all |x − p| < 1, we have
|x3 − p3 | =
Let δ = min
|(x − p)(x2 + xp + p2 )|
=
|x − p||x2 + xp + p2 |
≤
|x − p|[|x|2 + |x||p| + |p|2 ]
≤
|x − p|[[1 + |p|]2 + (1 + |p|)|p| + |p|2 ].
1,
[[1 + |p|]2 + (1 + |p|)|p| + |p|2 ]
|x3 − p3 |
> 0, then for all |x − p| < δ, we have
≤
|x − p|[[1 + |p|]2 + (1 + |p|)|p| + |p|2 ]
<
δ · [[1 + |p|]2 + (1 + |p|)|p| + |p|2 ]
≤
δ·
[[1 +
|p|]2
+ (1 + |p|)|p| + |p|2 ]
SOLUTION OF HW8
=
3
.
So we get
|x3 − p3 | < ,
for all |x − p| < δ.
That is, lim x3 = p3
x→p
4. [Problem 22, in Page 138] Let
f (x) =




 sin x,
if x ≤ c



 ax + b, if x > c.
Where a, b, c are constants. If b and c are given, and find all values of a for which f is continuous
at the point x = c.
Proof. By the definition of f , we know that
f (c+) = lim f (x) = lim [ax + b] = ac + b.
x&c
x&c
And
f (c−) = lim f (x) = lim sin x = sin c.
x%c
x&c
Since f is continuous at x = c, then f (c+) = f (c−), that is, we have
ac + b = sin c.
If c = 0, then we must have b = 0, that is, if b = c = 0, for all a ∈ R, then f is continuous at x = c.
If c 6= 0, then we have a =
sin c − b
.
c
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: [email protected]
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